Question: $w_1=2[\cos(\dfrac{\pi}{6})+i\sin(\dfrac{\pi}{6})]$ $w_2=17[\cos(\dfrac{5\pi}{12})+i\sin(\dfrac{5\pi}{12})]$ Express the product of $w_1$ and $w_2$ in polar form. The angle should be given in radians. $w_1\cdot w_2= $
Explanation: Background For any two complex numbers $z_1$ and $z_2$ (whose radii are $r_1$ and $r_2$ and angles are $\theta_1$ and $\theta_2$ ): The radius of $z_1\cdot z_2$ is the product of the original radii, $r_1 \cdot r_2$. The angle of $z_1\cdot z_2$ is the sum of the original angles, $\theta_1 + \theta_2$. In other words, suppose the polar forms of $z_1$ and $z_2$ are as follows, $z_1 = r_1[\cos(\theta_1) + {i}\sin(\theta_1)]$ $z_2 = r_2[\cos(\theta_2) + {i}\sin(\theta_2)]$, then the polar form of their product is: $z_1z_2 = r_1r_2[\cos(\theta_1 + \theta_2) + {i}\sin(\theta_1 + \theta_2)]$. [How do we get this?] Finding the radius of $w_1\cdot w_2$ $w_1=2[\cos(\dfrac{\pi}{6})+i\sin(\dfrac{\pi}{6})]$ $w_2=17[\cos(\dfrac{5\pi}{12})+i\sin(\dfrac{5\pi}{12})]$ Here, $r_1=2$ and $r_2=17$. Therefore, the radius of $w_1\cdot w_2$ is $r_1\cdot r_2=2\cdot17=34$. Finding the angle of $w_1\cdot w_2$ $w_1=2[\cos(\dfrac{\pi}{6})+i\sin(\dfrac{\pi}{6})]$ $w_2=17[\cos(\dfrac{5\pi}{12})+i\sin(\dfrac{5\pi}{12})]$ Here, $\theta_1=\dfrac{\pi}{6}$ and $\theta_2=\dfrac{5\pi}{12}$. Therefore, the angle of $w_1\cdot w_2$ is $\theta_1+\theta_2=\dfrac{\pi}{6}+\dfrac{5\pi}{12}=\dfrac{7\pi}{12}$. Summary We found that the radius of $w_1\cdot w_2$ is $34$ and its angle is $\dfrac{7\pi}{12}$. Therefore, $w_1\cdot w_2=34\left(\cos\left(\dfrac{7\pi}{12}\right)+i\sin\left(\dfrac{7\pi}{12}\right)\right)$